A 30 cm wrench is used to loosen a bolt with a force applied 0.3 mm from the bolt. It takes 60 N to loosen the bolt when the for

Answer:

10000    

Explanation:

When shifting the decimal point left, the number effectively gets divided by 10 for each movement.

Conversely, shifting it to the right corresponds to multiplying the number by 10 for every unit moved.

If the decimal point is moved four units to the left, it becomes 0.0003, which is equivalent to dividing 3.0 by 10000

Answer:

Explanation:

The length of a vector refers to its magnitude.

For a vector

R = a•i + b•j + c•k

The magnitude can be calculated using

|R|= √(a²+b²+c²)

Applying this formula to each given vector yields the following results.

(a) 2i + 4j + 3k

The length is

L = √(2²+4²+3²)

L = √(4+16+9)

L = √29

L = 5.385 unit

(b) 5i − 2j + k

Note that k represents 1k

The length is

L = √(5²+(-2)²+1²)

Because, -×- = +

L = √(25+4+1)

L = √30

L = 5.477 unit

(c) 2i − k

As there is no j component, it means that the j component is 0

L = 2i + 0j - 1k

The length is

L = √(2²+0²+(-1)²)

L = √(4+0+1)

L = √5

L = 2.236 unit

(d) 5i

Similarly, without a j-component and k-component

L = 5i + 0j + 0k

The length is

L = √(5²+0²+0²)

L = √(25+0+0)

L = √25

L = 5 unit

(e) 3i − 2j − k

The length is

L = √(3²+(-2)²+(-1)²)

L = √(9+4+1)

L = √14

L = 3.742 unit

(f) i + j + k

The length is

L = √(1²+1²+1²)

L = √(1+1+1)

L = √3

L = 1.7321 unit

Response:

a) 80 V

b) The electric field has a strength of 100 N/C, directed from point B toward point A, where the charge is negative.

Clarification:

Given:

An object with a charge of q = -6.00 x 10^-9 C starts from rest at point A, making its kinetic energy zero ( = 0) and moving to point B at a distance l = 0.500m where its kinetic energy is ( = 5.00 x 10^-7J). Additionally, the electric potential of q at point A is VA = +30.0 v.

Required:

(a) We seek to find the electric potential VB

(b) We need to compute the magnitude and orientation of the electric field E.

Solution

(a) Utilizing the given values for VA, and q, we derive a relationship among the three parameters and VB to compute VB.

At points A and B, the charge moves from A to B due to the electric field. The mechanical energy of the object remains conserved throughout this journey, allowing us to apply eq(1) in this context:

                                   .........................................(1)                                          

Where = 0, and the potential energy U of the charge is defined as U = q V

In this equation, V represents the electric potential. Thus, equation (1) can be expressed as:

                                  0+qVA= +qVB                    (Dividing by q)

                                         VA= /q + VB                  (Restructuring for VB)

                                         VB=VA- /q.......................................(2)

We now have the relation between VB, VA, and , allowing us to substitute our values for VA, , and q into equation (2) to obtain VB

                                         VB=VA- /q

                                              =30V-(5.00 x 10^-7J)/(-6.00 x 10^-9)

                                              =80 V

(b) After calculating VB, we may use equation a to derive the electric field E affecting the charge q, where the potential difference between the two points equals the integration of the electric field multiplied by the distance l between these points

                                   VA-VB =...................................(a)

                                               

                                   VA-VB=E                      (Restructuring for E)

                                            E= VA-VB/..................................(3)

Now, substituting our values for VA, Vs, and l into equation (3) allows us to compute the electric field E

                                            E= VA-VB/

                                              =-100 N/C

The electric field's magnitude equals 100 N/C and it directs from point B to point A towards the negative charge.

Answer:

The work performed by air resistance totals -0.0782 J

Explanation:

Hello!

According to the principle of conservation of energy, the energy of a raindrop must remain constant.

At the outset, the raindrop possesses only gravitational potential energy:

PE = m · g · h

Where:

PE = potential energy.

m = mass of the raindrop.

g = gravitational acceleration (9.8 m/s²)

h = height.

Let's determine the initial potential energy of the raindrop:

(4 mg should be converted into kg: 4 mg · 1 kg / 1 × 10⁶ mg = 4 × 10⁻⁶ kg)

PE = 4 × 10⁻⁶ kg · 9.8 m/s² · 2000 m

PE = 0.0784 J

As the raindrop descends, some of its potential energy converts into kinetic energy while the rest is lost to the air resistance. Upon reaching the ground, all initial potential energy has been either turned into kinetic energy or spent overcoming air resistance:

initial PE = final KE + Work by air

Where:

KE = kinetic energy.

Work by air = work done by air resistance.

The kinetic energy at ground level is computed as follows:

KE = 1/2 · m · v²

Where:

m = mass

v = velocity

<pThus:

KE = 1/2 · 4 × 10⁻⁶ kg · (10 m/s)²

KE = 2 × 10⁻⁴ J

Now, we can find the work done by air resistance:

initial PE = final KE + Work by air

0.0784 J = 2 × 10⁻⁴ J + Work by air

Work by air = 0.0784 J - 2 × 10⁻⁴ J

Work by air = 0.0782 J

Since work is performed in the opposite direction to movement, this results in a negative value. Therefore, the work done by air resistance is -0.0782 J.

Answer:

the temperature on the left side is 1.48 times greater than that on the right

Explanation:

GIVEN DATA:

T1 = 525 K

T2 = 275 K

It is known that

n and v are constant on both sides. Therefore we have

..............1

let the final pressure be P and the temperature

..................2

similarly

.............3

divide equation (2) by equation (3)

thus, the left side temperature equals 1.48 times the right side temperature